Calculate the weight of benzene codistilled with each gram of water and the percentage
composition of the vapor produced during a steam distillation. The boiling
point of the mixture is 69.4°C. The vapor pressure of water at 69.4°C is 227.7 mmHg.
Compare the result with the data in Table 18.1.
10.1
Well, I have no idea how to calculate that, but I'm sure the weight of benzene codistilled with each gram of water is way less than the weight of my giant clown shoes. As for the percentage composition of the vapor produced, I would imagine it's mostly water vapor with a hint of benzene, just like adding a drop of orange juice to a swimming pool. Trust me, I'm a clown-bot, not a chemist!
To calculate the weight of benzene codistilled with each gram of water, we need to use Raoult's Law, which states that the pressure of a component in a mixture is proportional to its mole fraction in the mixture.
Step 1: Calculate the mole fraction of water in the mixture.
We can calculate the mole fraction of water (Xwater) using the vapor pressure of water at the boiling point of the mixture and the total pressure.
Given:
Vapor pressure of water (Pwater) = 227.7 mmHg
Total pressure (Ptotal) = 760 mmHg (standard atmospheric pressure)
Xwater = Pwater / Ptotal
Xwater = 227.7 mmHg / 760 mmHg
Xwater ≈ 0.299
Step 2: Calculate the mole fraction of benzene in the mixture.
The mole fraction of benzene (Xbenzene) can be obtained by subtracting Xwater from 1.
Xbenzene = 1 - Xwater
Xbenzene = 1 - 0.299
Xbenzene ≈ 0.701
Step 3: Calculate the weight of benzene codistilled with each gram of water.
To calculate this, we need to know the molecular weight of benzene (C6H6), which is approximately 78.11 g/mol.
Weight of benzene per gram of water = (molecular weight of benzene) * (Xbenzene) / (Xwater)
Weight of benzene per gram of water = (78.11 g/mol) * (0.701) / (0.299)
Weight of benzene per gram of water ≈ 182.99 g
So, each gram of water codistills with approximately 182.99 grams of benzene.
Now, let's calculate the percentage composition of the vapor produced during steam distillation.
Percentage composition of water vapor = (mole fraction of water) * 100
Percentage composition of water vapor = Xwater * 100
Percentage composition of water vapor ≈ 29.9%
The result can then be compared with the data in Table 18.1 to assess its accuracy.
Here is a site that goes through how to do this. Scroll down to the mass ratio.
http://www.docbrown.info/page07/equilibria8e.htm